Question

The total energy of one mole of an ideal monatomic gas at 27°C is.
(A) 800 cal
(B) 900 cal
(C) 1000 cal
(D) 700 cal​

Answer 1

Answer:

(D) 900 cal ✔

EXPLANATION

➡️R = 8.314 j/mol.k

But answer is in calorie so the value of

R = 1.98 cal/mol .k

➡️ R = 2 cal/ mol.k (approx)

➡️T = 27 °c

T = 273 + 27 = 300k

NOw,

➡️Energy of one mole of an ideal monoatomic gas = 3/2 RT

$= \frac{3}{2} \times 2 \times 300$

= 900 cal

❇ HOPE IT HELPS YOU✨

Answer 2

Answer:

b 900cal......................